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Intersecting chord theorem:

14 * (2x+2)     =     12 * (2x+5)      solve for 'x'     then    you should be able to calculate the lengths and add them...

Image Here:

trapezoid area      1/2 (6 + 14)    *   13   = ........  m²

n = jenny now

k = jake now         five years ago   k-5 = 5(n-5)        k = 5n -20

   four years from now       k+4 = 2 (n+4)        k = 2n +4

Equate     5n-20   =  2n+4

                 3n = 24                               You can take it from here hopefully !   

-2 (-3x) + 5          simplify

I copied

"Triangle WXY has side lengths XY = 14 and WX = 8. The tangent to the circumcircle of triangle WXY at X is drawn, and the line through W that is parallel to this tangent intersects XY at Z. Find YZ."

into the web2 search box, it took me to google and I found the page straight away.

Use similar triangles

3/5 = (x+3)/20

Solve for x

x=9

4/5 = (y+4)/20

Solve for y

y=12

Do these 2 on your own ;)

3/5 = (x+u+3)/60         plug in x to get u

4/5 = (v+y+4)/60         plug in y to get v

That's it

V is given   as   490 pi

490 pi  = pi  r^2  h          sub in the value given for 'r'  and solve for 'h'   

                                            if no value is given for 'r' then your answer for 'h' will have an 'r' in it.....

(x-3-sqrt5)(x-3+sqrt5)

Multiply it out.....    then multiply by -2 to get it in the required form

then you will have  a  b   and  c      vertex will be at   x = - b/2a

   use this value of 'x' in the quadratic to calculate the 'y' component

Sorry, wrong answer!!

Guest #1

Your mistake is that you took 8,000 as the "Radius"  instead of 4,000 miles.

Simplify the L side

( 2 x⁹ y^3-b ) ^a         Which shows you that     2^a = 2^-3         a = -3      now you can try to find b  by equationg the 'y' exponents

y^3b-9  = y^{12      (take it from here)}

Simplify the following:
((4/(a^6))^3 (a^4 b)^3)/(12 a^4 (a^3 b^2)^4)

Multiply each exponent in a^4 b by 3:
((4/(a^6))^3 a^(3×4) b^3)/(12 a^4 (a^3 b^2)^4)

3×4 = 12:
((4/(a^6))^3 a^12 b^3)/(12 a^4 (a^3 b^2)^4)

Multiply each exponent in a^3 b^2 by 4:
((4/(a^6))^3 a^12 b^3)/(12 a^4 a^(4×3) b^(4×2))

4×2 = 8:
((4/(a^6))^3 a^12 b^3)/(12 a^4 a^(4×3) b^8)

4×3 = 12:
((4/(a^6))^3 a^12 b^3)/(12 a^4 a^12 b^8)

Multiply each exponent in 4/(a^6) by 3:
(4^3 a^(-6×3) a^12 b^3)/(12 a^4 a^12 b^8)

3 (-6) = -18:
(4^3 a^(-18) a^12 b^3)/(12 a^4 a^12 b^8)

4^3 = 4×4^2:
(4×4^2 a^12 b^3)/(a^18×12 a^4 a^12 b^8)

4^2 = 16:
(4×16 a^12 b^3)/(a^18×12 a^4 a^12 b^8)

4×16 = 64:
(64 a^12 b^3)/(a^18×12 a^4 a^12 b^8)

(64 a^12 b^3)/(a^18×12 a^4 a^12 b^8) = a^12/a^12×(64 b^3)/(a^18×12 a^4 b^8) = (64 b^3)/(a^18×12 a^4 b^8):
(64 b^3)/(a^18×12 a^4 b^8)

The gcd of 64 and 12 is 4, so (64 b^3)/(a^18×12 a^4 b^8) = ((4×16) b^3)/(a^18 (4×3) a^4 b^8) = 4/4×(16 b^3)/(a^18×3 a^4 b^8) = (16 b^3)/(a^18×3 a^4 b^8):
(16 b^3)/(a^18×3 a^4 b^8)

Combine powers. (16 b^3)/(a^18×3 a^4 b^8) = (16 a^(-4 - 18) b^(3 - 8))/3:
(16 a^(-4 - 18) b^(3 - 8))/3

-4 - 18 = -22:
(16 a^(-22) b^(3 - 8))/3

3 - 8 = -5:

 (16 b^(-5)) / (3 a^22)

Notice that x-2 and x+1 are factors of x^2-x-2.

\(\frac{x+7}{x^2-x-2}=\frac{a}{x-2}+\frac{b}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{a(x+1)}{x^2-x-2}+\frac{b(x-2)}{x^2-x-2}\\ x-7=a(x+1)+b(x-2)\\ x-7=ax+a+bx-2b\\\)

ax+bx must add up to the x term on the left side, which is x. a-2b must subtract to the constant on the left side, -7. We then have a system of equations.

a+b=1

a-2b=-7

We find that a= -5/3 and b=8/3.

There is a very simple and quite accurate formula that can be used to meassure the distance to the Horizon:

D =Sqrt[1.5  x  H],  where H = Height in feet and D = The distance in Miles.

a - Sqrt[1.5 x 5,280] =~89 miles

b - Sqrt[1.5 x  31,680] =~218 miles.

c - Sqrt[1.5 x  528,000] =~890 miles.

The number of paths of length 2 is (1 + 1 + 2 + 3 + 3 + 3 + 4 + 4 + 5 + 8)/2 = 34/2 = 17.

Thanks much. Appreciate decent people 

From the formula for a geometric series, the sum is (3/2)/(1 - 4/9) = 27/10.

The sum of the coordinates is 3 + 8 = 11.

We  can  figure the distance  to the horizon  using a right triangle

The distance from  the center of the Earth to any object above the Earth's surface  = the hypotenuse = 8000  + h      where h is the height of the object [in  miles ] above the Earth's surface

And the radius of the Earth  is equal to one leg of the right triangle

And the distance from the object to the horizon, D,  is the other leg

So...we have this general formula to determine D

D  = √ [ (8000 + h)^2  - 8000^2]

So....

(a)  D  = √ [ (8000 + 1)^2  - 8000^2 ]  ≈ 126.5 miles

(b)  D  =   √ [ (8000 + 6)^2  - 8000^2 ]  ≈  309.9  miles

(c)  D  = √ [ (8000 + 100)^2  - 8000^2 ]  ≈  1268.9 miles

The ratio of the second term to the first term is 8/3.